CHAPTER NINETEEN: PROCESSES COROUTINES AND CONCURRENCY (Part 5)

The Art of ASSEMBLY LANGUAGE PROGRAMMING

Chapter Nineteen (Part 4)	Table of Content	Chapter Nineteen (Part 6)

CHAPTER NINETEEN:
PROCESSES COROUTINES AND CONCURRENCY (Part 5)

19.2.2 - Dynamic Shared Memory

19.2.2 Dynamic Shared Memory

Although the static shared memory the previous section describes is very useful it does suffer from a few limitations. First of all any program that uses the global shared segment must be aware of the location of every other program that uses the shared segment. This effectively means that the use of the shared segment is limited to a single set of cooperating processes at any one given time. You cannot have two independent sets of programs using the shared memory at the same time. Another limitation with the static system is that you must know the size of all variables when you write your program you cannot create dynamic data structures whose size varies at run time. It would be nice for example to have calls like shmalloc and shmfree that let you dynamically allocate and free memory in a shared region. Fortunately it is very easy to overcome these limitations by creating a dynamic shared memory manager.

A reasonable shared memory manager will have four functions: initialize shmalloc shmattach and shmfree. The initialization call reclaims all shared memory in use. The shmalloc call lets a process allocate a new block of shared memory. Only one process in a group of cooperating processes makes this call. Once shmalloc allocates a block of memory the other processes use the shmattach call to obtain the address of the shared memory block. The following code implements a dynamic shared memory manager. The code is similar to that appearing in the Standard Library except this code allows a maximum of 64K storage on the heap.

; SHMALLOC.ASM ; ; This TSR sets up a dynamic shared memory system. ; ; This TSR checks to make sure there isn't a copy already active in ; memory. When removing itself from memory it makes sure there are ; no other interrupts chained into INT 2Fh before doing the remove. ; ; ; ; The following segments must appear in this order and before the ; Standard Library includes. ResidentSeg segment para public 'Resident' ResidentSeg ends SharedMemory segment para public 'Shared' SharedMemory ends EndResident segment para public 'EndRes' EndResident ends .xlist .286 include stdlib.a includelib stdlib.lib .list ; Resident segment that holds the TSR code: ResidentSeg segment para public 'Resident' assume cs:ResidentSeg ds:nothing NULL equ 0 ; Data structure for an allocated data region. ; ; Key- user supplied ID to associate this region with a particular set ; of processes. ; ; Next- Points at the next allocated block. ; Prev- Points at the previous allocated block. ; Size- Size (in bytes) of allocated block not including header structure. Region struct key word ? next word ? prev word ? blksize word ? Region ends Startmem equ Region ptr [0] AllocatedList word 0 ;Points at chain of alloc'd blocks. FreeList word 0 ;Points at chain of free blocks. ; Int 2Fh ID number for this TSR: MyTSRID byte 0 byte 0 ;Padding so we can print it. ; PSP is the psp address for this program. PSP word 0 OldInt2F dword ? ; MyInt2F- Provides int 2Fh (multiplex interrupt) support for this ; TSR. The multiplex interrupt recognizes the following ; subfunctions (passed in AL): ; ; 00h- Verify presence. Returns 0FFh in AL and a pointer ; to an ID string in es:di if the ; TSR ID (in AH) matches this ; particular TSR. ; ; 01h- Remove. Removes the TSR from memory. ; Returns 0 in AL if successful ; 1 in AL if failure. ; ; 11h- shmalloc CX contains the size of the block ; to allocate. ; DX contains the key for this block. ; Returns a pointer to block in ES:DI ; and size of allocated block in CX. ; Returns an error code in AX. Zero ; is no error one is "key already ; exists " two is "insufficient ; memory for request." ; ; 12h- shmfree DX contains the key for this block. ; This call frees the specified block ; from memory. ; ; 13h- shminit Initializes the shared memory system ; freeing all blocks currently in ; use. ; ; 14h- shmattach DX contains the key for a block. ; Search for that block and return ; its address in ES:DI. AX contains ; zero if successful three if it ; cannot locate a block with the ; specified key. MyInt2F proc far assume ds:nothing cmp ah MyTSRID ;Match our TSR identifier? je YepItsOurs jmp OldInt2F ; Okay we know this is our ID now check for a verify remove or ; return segment call. YepItsOurs: cmp al 0 ;Verify Call jne TryRmv mov al 0ffh ;Return success. lesi IDString iret ;Return back to caller. IDString byte "Dynamic Shared Memory TSR" 0 TryRmv: cmp al 1 ;Remove call. jne Tryshmalloc ; See if we can remove this TSR: push es mov ax 0 mov es ax cmp word ptr es:[2Fh*4] offset MyInt2F jne TRDone cmp word ptr es:[2Fh*4 + 2] seg MyInt2F je CanRemove ;Branch if we can. TRDone: mov ax 1 ;Return failure for now. pop es iret ; Okay they want to remove this guy *and* we can remove it from memory. ; Take care of all that here. assume ds:ResidentSeg CanRemove: push ds pusha cli ;Turn off the interrupts while mov ax 0 ; we mess with the interrupt mov es ax ; vectors. mov ax cs mov ds ax mov ax word ptr OldInt2F mov es:[2Fh*4] ax mov ax word ptr OldInt2F+2 mov es:[2Fh*4 + 2] ax ; Okay one last thing before we quit- Let's give the memory allocated ; to this TSR back to DOS. mov ds PSP mov es ds:[2Ch] ;Ptr to environment block. mov ah 49h ;DOS release memory call. int 21h mov ax ds ;Release program code space. mov es ax mov ah 49h int 21h popa pop ds pop es mov ax 0 ;Return Success. iret ; Stick BadKey here so that it is close to its associated branch (from below). ; ; If come here we've discovered an allocated block with the ; specified key. Return an error code (AX=1) and the size of that ; allocated block (in CX). BadKey: mov cx [bx].Region.BlkSize mov ax 1 ;Already allocated error. pop bx pop ds iret ; See if this is a shmalloc call. ; If so on entry - ; DX contains the key. ; CX contains the number of bytes to allocate. ; ; On exit: ; ; ES:DI points at the allocated block (if successful). ; CX contains the actual size of the allocated block (>=CX on entry). ; AX contains error code 0 if no error. Tryshmalloc: cmp al 11h ;shmalloc function code. jne Tryshmfree ; First search through the allocated list to see if a block with the ; current key number already exists. DX contains the requested key. assume ds:SharedMemory assume bx:ptr Region assume di:ptr Region push ds push bx mov bx SharedMemory mov ds bx mov bx ResidentSeg:AllocatedList test bx bx ;Anything on this list? je SrchFreeList SearchLoop: cmp dx [bx].Key ;Key exist already? je BadKey mov bx [bx].Next ;Get next region. test bx bx ;NULL? if not try another jne SearchLoop ; entry in the list. ; If an allocated block with the specified key does not already exist ; then try to allocate one from the free memory list. SrchFreeList: mov bx ResidentSeg:FreeList test bx bx ;Empty free list? je OutaMemory FirstFitLp: cmp cx [bx].BlkSize ;Is this block big enough? jbe GotBlock mov bx [bx].Next ;If not on to the next one. test bx bx ;Anything on this list? jne FirstFitLp ; If we drop down here we were unable to find a block that was large ; enough to satisfy the request. Return an appropriate error OutaMemory: mov cx 0 ;Nothing available. mov ax 2 ;Insufficient memory error. pop bx pop ds iret ; If we find a large enough block we've got to carve the new block ; out of it and return the rest of the storage to the free list. If the ; free block is at least 32 bytes larger than the requested size we will ; do this. If the free block is less than 32 bytes larger we will simply ; give this free block to the requesting process. The reason for the ; 32 bytes is simple: We need eight bytes for the new block's header ; (the free block already has one) and it doesn't make sense to fragment ; blocks to sizes below 24 bytes. That would only increase processing time ; when processes free up blocks by requiring more work coalescing blocks. GotBlock: mov ax [bx].BlkSize ;Compute difference in size. sub ax cx cmp ax 32 ;At least 32 bytes left? jbe GrabWholeBlk ;If not take this block. ; Okay the free block is larger than the requested size by more than 32 ; bytes. Carve the new block from the end of the free block (that way ; we do not have to change the free block's pointers only the size. mov di bx add di [bx].BlkSize ;Scoot to end minus 8 sub di cx ;Point at new block. sub [bx].BlkSize cx ;Remove alloc'd block and sub [bx].BlkSize 8 ; room for header. mov [di].BlkSize cx ;Save size of block. mov [di].Key dx ;Save key. ; Link the new block into the list of allocated blocks. mov bx ResidentSeg:AllocatedList mov [di].Next bx mov [di].Prev NULL ;NULL previous pointer. test bx bx ;See if it was an empty list. je NoPrev mov [bx].Prev di ;Set prev ptr for old guy. NoPrev: mov ResidentSeg:AllocatedList di RmvDone: add di 8 ;Point at actual data area. mov ax ds ;Return ptr in es:di. mov es ax mov ax 0 ;Return success. pop bx pop ds iret ; If the current free block is larger than the request but not by more ; that 32 bytes just give the whole block to the user. GrabWholeBlk: mov di bx mov cx [bx].BlkSize ;Return actual size. cmp [bx].Prev NULL ;First guy in list? je Rmv1st cmp [bx].Next NULL ;Last guy in list? je RmvLast ; Okay this record is sandwiched between two other in the free list. ; Cut it out from among the two. mov ax [bx].Next ;Save the ptr to the next mov bx [bx].Prev ; item in the prev item's mov [bx].Next ax ; next field. mov ax bx ;Save the ptr to the prev mov bx [di].Next ; item in the next item's mov [bx].Prev bx ; prev field. jmp RmvDone ; The block we want to remove is at the beginning of the free list. ; It could also be the only item on the free list! Rmv1st: mov ax [bx].Next mov FreeList ax ;Remove from free list. jmp RmvDone ; If the block we want to remove is at the end of the list handle that ; down here. RmvLast: mov bx [bx].Prev mov [bx].Next NULL jmp RmvDone assume ds:nothing bx:nothing di:nothing ; This code handles the SHMFREE function. ; On entry DX contains the key for the block to free. We need to ; search through the allocated block list and find the block with that ; key. If we do not find such a block this code returns without doing ; anything. If we find the block we need to add its memory to the ; free pool. However we cannot simply insert this block on the front ; of the free list (as we did for the allocated blocks). It might ; turn out that this block we're freeing is adjacent to one or two ; other free blocks. This code has to coalesce such blocks into ; a single free block. Tryshmfree: cmp al 12h jne Tryshminit ; First search the allocated block list to see if we can find the ; block to remove. If we don't find it in the list anywhere just return. assume ds:SharedMemory assume bx:ptr Region assume di:ptr Region push ds push di push bx mov bx SharedMemory mov ds bx mov bx ResidentSeg:AllocatedList test bx bx ;Empty allocated list? je FreeDone SrchList: cmp dx [bx].Key ;Search for key in DX. je FoundIt mov bx [bx].Next test bx bx ;At end of list? jne SrchList FreeDone: pop bx pop di ;Nothing allocated just pop ds ; return to caller. iret ; Okay we found the block the user wants to delete. Remove it from ; the allocated list. There are three cases to consider: ; (1) it is at the front of the allocated list (2) it is at the end of ; the allocated list and (3) it is in the middle of the allocated list. FoundIt: cmp [bx].Prev NULL ;1st item in list? je Free1st cmp [bx].Next NULL ;Last item in list? je FreeLast ; Okay we're removing an allocated item from the middle of the allocated ; list. mov di [bx].Next ;[next].prev := [cur].prev mov ax [bx].Prev mov [di].Prev ax xchg ax di mov [di].Next ax ;[prev].next := [cur].next jmp AddFree ; Handle the case where we are removing the first item from the allocation ; list. It is possible that this is the only item on the list (i.e. it ; is the first and last item) but this code handles that case without any ; problems. Free1st: mov ax [bx].Next mov ResidentSeg:AllocatedList ax jmp AddFree ; If we're removing the last guy in the chain simply set the next field ; of the previous node in the list to NULL. FreeLast: mov di [bx].Prev mov [di].Next NULL ; Okay now we've got to put the freed block onto the free block list. ; The free block list is sorted according to address. We have to search ; for the first free block whose address is greater than the block we've ; just freed and insert the new free block before that one. If the two ; blocks are adjacent then we've got to merge them into a single free ; block. Also if the block before is adjacent we must merge it as ; well. This will coalesce all free blocks on the free list so there ; are as few free blocks as possible and those blocks are as large as ; possible. AddFree: mov ax ResidentSeg:FreeList test ax ax ;Empty list? jne SrchPosn ; If the list is empty stick this guy on as the only entry. mov ResidentSeg:FreeList bx mov [bx].Next NULL mov [bx].Prev NULL jmp FreeDone ; If the free list is not empty search for the position of this block ; in the free list: SrchPosn: mov di ax cmp bx di jb FoundPosn mov ax [di].Next test ax ax ;At end of list? jne SrchPosn ; If we fall down here the free block belongs at the end of the list. ; See if we need to merge the new block with the old one. mov ax di add ax [di].BlkSize ;Compute address of 1st byte add ax 8 ; after this block. cmp ax bx je MergeLast ; Okay just add the free block to the end of the list. mov [di].Next bx mov [bx].Prev di mov [bx].Next NULL jmp FreeDone ; Merge the freed block with the block DI points at. MergeLast: mov ax [di].BlkSize add ax [bx].BlkSize add ax 8 mov [di].BlkSize ax jmp FreeDone ; If we found a free block before which we are supposed to insert ; the current free block drop down here and handle it. FoundPosn: mov ax bx ;Compute the address of the add ax [bx].BlkSize ; next block in memory. add ax 8 cmp ax di ;Equal to this block? jne DontMerge ; The next free block is adjacent to the one we're freeing so just ; merge the two. mov ax [di].BlkSize ;Merge the sizes together. add ax 8 add [bx].BlkSize ax mov ax [di].Next ;Tweak the links. mov [bx].Next ax mov ax [di].Prev mov [bx].Prev ax jmp TryMergeB4 ; If the blocks are not adjacent just link them together here. DontMerge: mov ax [di].Prev mov [di].Prev bx mov [bx].Prev ax mov [bx].Next di ; Now see if we can merge the current free block with the previous free blk. TryMergeB4: mov di [bx].Prev mov ax di add ax [di].BlkSize add ax 8 cmp ax bx je CanMerge pop bx pop di ;Nothing allocated just pop ds ; return to caller. iret ; If we can merge the previous and current free blocks do that here: CanMerge: mov ax [bx].Next mov [di].Next ax mov ax [bx].BlkSize add ax 8 add [di].BlkSize ax pop bx pop di pop ds iret assume ds:nothing assume bx:nothing assume di:nothing ; Here's where we handle the shared memory initializatin (SHMINIT) function. ; All we got to do is create a single block on the free list (which is all ; available memory) empty out the allocated list and then zero out all ; shared memory. Tryshminit: cmp al 13h jne TryShmAttach ; Reset the memory allocation area to contain a single free block of ; memory whose size is 0FFF8h (need to reserve eight bytes for the block's ; data structure). push es push di push cx mov ax SharedMemory ;Zero out the shared mov es ax ; memory segment. mov cx 32768 xor ax ax mov di ax rep stosw ; Note: the commented out lines below are unnecessary since the code above ; has already zeroed out the entire shared memory segment. ; Note: we cannot put the first record at offset zero because offset zero ; is the special value for the NULL pointer. We'll use 4 instead. mov di 4 ; mov es:[di].Region.Key 0 ;Key is arbitrary. ; mov es:[di].Region.Next 0 ;No other entries. ; mov es:[di].Region.Prev 0 ; Ditto. mov es:[di].Region.BlkSize 0FFF8h ;Rest of segment. mov ResidentSeg:FreeList di pop cx pop di pop es mov ax 0 ;Return no error. iret ; Handle the SHMATTACH function here. On entry DX contains a key number. ; Search for an allocated block with that key number and return a pointer ; to that block (if found) in ES:DI. Return an error code (AX=3) if we ; cannot find the block. TryShmAttach: cmp al 14h ;Attach opcode. jne IllegalOp mov ax SharedMemory mov es ax mov di ResidentSeg:AllocatedList FindOurs: cmp dx es:[di].Region.Key je FoundOurs mov di es:[di].Region.Next test di di jne FoundOurs mov ax 3 ;Can't find the key. iret FoundOurs: add di 8 ;Point at actual data. mov ax 0 ;No error. iret ; They called us with an illegal subfunction value. Try to do as little ; damage as possible. IllegalOp: mov ax 0 ;Who knows what they were thinking? iret MyInt2F endp assume ds:nothing ResidentSeg ends ; Here's the segment that will actually hold the shared data. SharedMemory segment para public 'Shared' db 0FFFFh dup (?) SharedMemory ends cseg segment para public 'code' assume cs:cseg ds:ResidentSeg ; SeeIfPresent- Checks to see if our TSR is already present in memory. ; Sets the zero flag if it is clears the zero flag if ; it is not. SeeIfPresent proc near push es push ds push di mov cx 0ffh ;Start with ID 0FFh. IDLoop: mov ah cl push cx mov al 0 ;Verify presence call. int 2Fh pop cx cmp al 0 ;Present in memory? je TryNext strcmpl byte "Dynamic Shared Memory TSR" 0 je Success TryNext: dec cl ;Test USER IDs of 80h..FFh js IDLoop cmp cx 0 ;Clear zero flag. Success: pop di pop ds pop es ret SeeIfPresent endp ; FindID- Determines the first (well last actually) TSR ID available ; in the multiplex interrupt chain. Returns this value in ; the CL register. ; ; Returns the zero flag set if it locates an empty slot. ; Returns the zero flag clear if failure. FindID proc near push es push ds push di mov cx 0ffh ;Start with ID 0FFh. IDLoop: mov ah cl push cx mov al 0 ;Verify presence call. int 2Fh pop cx cmp al 0 ;Present in memory? je Success dec cl ;Test USER IDs of 80h..FFh js IDLoop xor cx cx cmp cx 1 ;Clear zero flag Success: pop di pop ds pop es ret FindID endp Main proc meminit mov ax ResidentSeg mov ds ax mov ah 62h ;Get this program's PSP int 21h ; value. mov PSP bx ; Before we do anything else we need to check the command line ; parameters. If there is one and it is the word "REMOVE" then remove ; the resident copy from memory using the multiplex (2Fh) interrupt. argc cmp cx 1 ;Must have 0 or 1 parms. jb TstPresent je DoRemove Usage: print byte "Usage:" cr lf byte " shmalloc" cr lf byte "or shmalloc REMOVE" cr lf 0 ExitPgm ; Check for the REMOVE command. DoRemove: mov ax 1 argv stricmpl byte "REMOVE" 0 jne Usage call SeeIfPresent je RemoveIt print byte "TSR is not present in memory cannot remove" byte cr lf 0 ExitPgm RemoveIt: mov MyTSRID cl printf byte "Removing TSR (ID #%d) from memory..." 0 dword MyTSRID mov ah cl mov al 1 ;Remove cmd ah contains ID int 2Fh cmp al 1 ;Succeed? je RmvFailure print byte "removed." cr lf 0 ExitPgm RmvFailure: print byte cr lf byte "Could not remove TSR from memory." cr lf byte "Try removing other TSRs in the reverse order " byte "you installed them." cr lf 0 ExitPgm ; Okay see if the TSR is already in memory. If so abort the ; installation process. TstPresent: call SeeIfPresent jne GetTSRID print byte "TSR is already present in memory." cr lf byte "Aborting installation process" cr lf 0 ExitPgm ; Get an ID for our TSR and save it away. GetTSRID: call FindID je GetFileName print byte "Too many resident TSRs cannot install" cr lf 0 ExitPgm ; Things look cool so far so install the interrupts GetFileName: mov MyTSRID cl print byte "Installing interrupts..." 0 ; Patch into the INT 2Fh interrupt chain. cli ;Turn off interrupts! mov ax 0 mov es ax mov ax es:[2Fh*4] mov word ptr OldInt2F ax mov ax es:[2Fh*4 + 2] mov word ptr OldInt2F+2 ax mov es:[2Fh*4] offset MyInt2F mov es:[2Fh*4+2] seg ResidentSeg sti ;Okay ints back on. ; We're hooked up the only thing that remains is to initialize the shared ; memory segment and then terminate and stay resident. printf byte "Installed TSR ID #%d." cr lf 0 dword MyTSRID mov ah MyTSRID ;Initialization call. mov al 13h int 2Fh mov dx EndResident ;Compute size of program. sub dx PSP mov ax 3100h ;DOS TSR command. int 21h Main endp cseg ends sseg segment para stack 'stack' stk db 256 dup (?) sseg ends zzzzzzseg segment para public 'zzzzzz' LastBytes db 16 dup (?) zzzzzzseg ends end Main

We can modify the two applications from the previous section to try out this code:

; SHMAPP3.ASM ; ; This is a shared memory application that uses the dynamic shared memory ; TSR (SHMALLOC.ASM). This program inputs a string from the user and ; passes that string to SHMAPP4.ASM through the shared memory area. ; ; .xlist include stdlib.a includelib stdlib.lib .list dseg segment para public 'data' ShmID byte 0 dseg ends cseg segment para public 'code' assume cs:cseg ds:dseg es:SharedMemory ; SeeIfPresent- Checks to see if the shared memory TSR is present in memory. ; Sets the zero flag if it is clears the zero flag if ; it is not. This routine also returns the TSR ID in CL. SeeIfPresent proc near push es push ds push di mov cx 0ffh ;Start with ID 0FFh. IDLoop: mov ah cl push cx mov al 0 ;Verify presence call. int 2Fh pop cx cmp al 0 ;Present in memory? je TryNext strcmpl byte "Dynamic Shared Memory TSR" 0 je Success TryNext: dec cl ;Test USER IDs of 80h..FFh js IDLoop cmp cx 0 ;Clear zero flag. Success: pop di pop ds pop es ret SeeIfPresent endp ; The main program for application #1 links with the shared memory ; TSR and then reads a string from the user (storing the string into ; shared memory) and then terminates. Main proc assume cs:cseg ds:dseg es:SharedMemory mov ax dseg mov ds ax meminit print byte "Shared memory application #3" cr lf 0 ; See if the shared memory TSR is around: call SeeIfPresent je ItsThere print byte "Shared Memory TSR (SHMALLOC) is not loaded." cr lf byte "This program cannot continue execution." cr lf 0 ExitPgm ; Get the input line from the user: ItsThere: mov ShmID cl print byte "Enter a string: " 0 lea di InputLine ;ES already points at proper seg. getsm ; The string is in our heap space. Let's move it over to the shared ; memory segment. strlen inc cx ;Add one for zero byte. push es push di mov dx 1234h ;Our "key" value. mov ah ShmID mov al 11h ;Shmalloc call. int 2Fh mov si di ;Save as dest ptr. mov dx es pop di ;Retrive source address. pop es strcpy ;Copy from local to shared. print byte "Entered '" 0 puts print byte "' into shared memory." cr lf 0 Quit: ExitPgm ;DOS macro to quit program. Main endp cseg ends sseg segment para stack 'stack' stk db 1024 dup ("stack ") sseg ends zzzzzzseg segment para public 'zzzzzz' LastBytes db 16 dup (?) zzzzzzseg ends end Main

; SHMAPP4.ASM ; ; This is a shared memory application that uses the dynamic shared memory ; TSR (SHMALLOC.ASM). This program assumes the user has already run the ; SHMAPP3 program to insert a string into shared memory. This program ; simply prints that string from shared memory. ; .xlist include stdlib.a includelib stdlib.lib .list dseg segment para public 'data' ShmID byte 0 dseg ends cseg segment para public 'code' assume cs:cseg ds:dseg es:SharedMemory ; SeeIfPresent- Checks to see if the shared memory TSR is present in memory. ; Sets the zero flag if it is clears the zero flag if ; it is not. This routine also returns the TSR ID in CL. SeeIfPresent proc near push es push ds push di mov cx 0ffh ;Start with ID 0FFh. IDLoop: mov ah cl push cx mov al 0 ;Verify presence call. int 2Fh pop cx cmp al 0 ;Present in memory? je TryNext strcmpl byte "Dynamic Shared Memory TSR" 0 je Success TryNext: dec cl ;Test USER IDs of 80h..FFh js IDLoop cmp cx 0 ;Clear zero flag. Success: pop di pop ds pop es ret SeeIfPresent endp ; The main program for application #1 links with the shared memory ; TSR and then reads a string from the user (storing the string into ; shared memory) and then terminates. Main proc assume cs:cseg ds:dseg es:SharedMemory mov ax dseg mov ds ax meminit print byte "Shared memory application #4" cr lf 0 ; See if the shared memory TSR is around: call SeeIfPresent je ItsThere print byte "Shared Memory TSR (SHMALLOC) is not loaded." cr lf byte "This program cannot continue execution." cr lf 0 ExitPgm ; If the shared memory TSR is present get the address of the shared segment ; into the ES register: ItsThere: mov ah cl ;ID of our TSR. mov al 14h ;Attach call mov dx 1234h ;Our "key" value int 2Fh ; Print the string input in SHMAPP3: print byte "String from SHMAPP3 is '" 0 puts print byte "' from shared memory." cr lf 0 Quit: ExitPgm ;DOS macro to quit program. Main endp cseg ends sseg segment para stack 'stack' stk db 1024 dup ("stack ") sseg ends zzzzzzseg segment para public 'zzzzzz' LastBytes db 16 dup (?) zzzzzzseg ends end Main